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4 years ago

How is an insoluble solid separated from a liquid in a mixture?

Chemistry

1 answer:

abruzzese [7]4 years ago

7 0

Answer:

C = through filter

Explanation:

If the particle size is is greater than 1000 nm then particles can be separated through the filtration. These particles are large enough and can be seen through naked eye.

If the particles size is between 1 - 1000 nm i.e, 0.001- 1μm then it can not be separated through filtration. The pore size of filter paper is 2μm and particles pas through it. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

$m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3$

Best regards!

5 0

3 years ago

Millimeters, centimeters, meters, kilometers, Inches,<br> feet, and miles are all examples of

siniylev [52]

Answer:

Distance, some kind of distance or length.

Explanation:

4 0

3 years ago

In which way does a na1+ ion differ from a neutral na atom?

DochEvi [55]

Na 1+ misses 1 electron

8 0

4 years ago

In the figure below, two nonpolar molecules are interacting.Which interaction would most likely cause these molecules to repel e

Fed [463]

Two non-polar molecules are most likely to interact by
induced dipole-induced dipole interaction.

Non-polar substances do not have a permanently established charge distribution due to similar electron affinities of the atoms that are present. Moreover, due to the absence of a polar hydrogen, they cannot exhibit hydrogen bonding. They interact with one another by induced dipole-induced dipole interactions which arise from the molecules of the substances coming into close vicinity of one another.

6 0

3 years ago

Read 2 more answers

What is the formula of a compound in which element y form ccp lattice and atom x occupy 1/3rd of tetrahedral voids

Kisachek [45]

Y : CCP : 4 atoms
X : tetrahedral voids would be 1/3 × 8 = 8/3

so formula would be Y12X8 or Y3X2 !!

5 0

4 years ago